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by nlpfollower » Mon Mar 15, 2010 1:47 am
Image

The outline of a sign for an ice-cream store is made by placing 3/4 of the circumference of a circle with radius 2 feet on top of an isosceles triangle with height 5 feet, as shown above. What is the perimeter, in feet, of the sign?

A. 3pi + 3√3
B. 3pi + 6√3
A. 3pi + 2√33
A. 4pi + 3√3
A. 4pi + 6√3

The OA is B. Please give an explanation on how the perimeter of the triangular part has been calculated.
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by kstv » Mon Mar 15, 2010 2:45 am
3/4 of Circumference = 2pi r*3/4 = 3pi
The chord which is the base of the isosceles traingle will subtend an right angle at the centre
its length will be 2√ 2
slant side of the cone is h²+ (base/2)²
= 25 + 2 = √ 27
perimeter of 2 slant side = 2*√ 27 = 6√ 3
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by [email protected] » Mon Mar 15, 2010 2:57 am
kstv wrote: The chord which is the base of the isosceles traingle will subtend an right angle at the centre
its length will be 2√ 2

kstv. Can you explain how the base of the triangle is of length 2√ 2 ???
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by kstv » Mon Mar 15, 2010 6:14 am
In the figure attached. The green line is the chord. The blue lines are the radius measuring 2 units. The arc that the chord makes is 1/4 of the circumference i.e. 1/4 of 360° so it is a right angle. The chord is the hypotenuse = √ r² + r²
r = 2 so the hypotenuse is √ 8 = 2√ 2
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by gmat bull » Mon Mar 15, 2010 11:14 am
Can somebody please explain the flaw in my analysis of this problem?

Since we know that this is a 30-60-90 triangle, we can find sides by applying x-x√3-2x ratio.

x√3=5
x= 5√3/3

So hypotenuse of the 30-60-90 triangle is 10√3/3
And perimeter of two sides is 10√3/3*2

Where did I make a mistake?
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by gmat bull » Mon Mar 15, 2010 11:15 am
Can somebody please explain the flaw in my analysis of this problem?

Since we know that this is a 30-60-90 triangle, we can find sides by applying x-x√3-2x ratio.

x√3=5
x= 5√3/3

So hypotenuse of the 30-60-90 triangle is 10√3/3
And perimeter of two sides is 10√3/3*2

Where did I make a mistake?
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by kstv » Wed Mar 17, 2010 3:25 am
It is an isosceles triangle. So the height will bisect the base or the chord. If the resultant triangle is a it is a 30 60 90
it means that the angle at the vertex is 60. Then it ia a equilateral triangle. Not a valid assumption.
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by treker » Wed Mar 17, 2010 7:32 pm
kstv wrote:3/4 of Circumference = 2pi r*3/4 = 3pi
The chord which is the base of the isosceles traingle will subtend an right angle at the centre
its length will be 2√ 2
slant side of the cone is h²+ (base/2)²
= 25 + 2 = √ 27
perimeter of 2 slant side = 2*√ 27 = 6√ 3
How did you know that the chord makes an angle of 90 deg at the center of the circle? Thanks!
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by analyst218 » Thu Mar 18, 2010 12:52 pm
treker wrote:
kstv wrote:3/4 of Circumference = 2pi r*3/4 = 3pi
The chord which is the base of the isosceles traingle will subtend an right angle at the centre
its length will be 2√ 2
slant side of the cone is h²+ (base/2)²
= 25 + 2 = √ 27
perimeter of 2 slant side = 2*√ 27 = 6√ 3
How did you know that the chord makes an angle of 90 deg at the center of the circle? Thanks!
because we know the circumference up to the chord is 3/4. thus the arc with the two pts of the chord is 1/4.
360/4 = 90.
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